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How come a turboprop can't reach speeds as high as a turbofan?
I understand that turbofan powered aircraft can reach much faster speeds than turboprops like the C130 Hercules, and I can't imagine a turbofan turboprop powered 737 would be of any use really, but as to the logic behind that, I'm a bit confused.
I've read up that a turboprop draws in more air but at a slower speed and a turbofan draws in less air but at a higher speed. But if you have more air in a turboprop, can't you compress it to great amounts, whereas in a turbofan you have less air so you cannot compress as much? And to my understanding (and I may be wrong) the low pressure compressor in a turbofan regulates not only the pressure but the speed of the air, so how is the entry speed of the air relevant to the engine's performance? It just seems a bit counter-intuitive to me. More air = more to compress, but lower speed = less being sucked into the engine per second?
Somewhere online it said that turboprops aren't great at high altitudes unlike a turbofan, because the air gets thinner and the turboprop can't suck as much in then. But then my question is, if a turbofan sucks in less then how does this statement work?
I'm just completely stuck with this at the moment, and I've had a bit of a mindblock. If someone could help me out with this I'd really appreciate it.

"I can't imagine a turbofan powered 737 would be of any use really" Shhh.... don't tell Southwest Airlines - their entire fleet of aircraft are turbofan-powered 737s.

Props must spin very fast to maintain optimal AoA at higher IAS.  Once they go supersonic, drag ruins advantages they have over fans as "spinning high Aspect Ratio wings".  However, higher airspeeds have been successfully achieved with the Tu-95 contrarotating turbo-prop powered by 2 jet cores.  Makes one wonder if that's what they originally had in mind for the B-52.

A turboprop 737 would have its blades hit the ground.  Engines on 737 are really low.  Boeing really painted themselves into a corner by not putting tall gear on the 737 Classic.

"I can't imagine a turbofan powered 737 would be of any use really" Did you mean to say "turboprop" here? All 737s ever built have been powered by turbofans (though the -100s and -200s were powered by low-bypass turbofans rather than the more modern high-bypass ones.)

Supersonic propellers are also unbelievably loud (the XF-84H was known as the "thunderscreech" because of this, allegedly audible 40 km away even when it was on the ground), which is not a great attribute for airliners

"Somewhere online it said that". Where would that be? Citation please.

@Harper-ReinstateMonica You might could do it, by mounting the engines above the wings, the way the P-3 does. Which would be, congratulations, you just reinvented the Lockheed Electra! Which did certain thing pretty well, but hardly an efficient replacement for a 737.

The attributes of the incoming air for combustion matter less than how the thrust is generated.  Many turbofans operate past Mach 1, but a turboprop's thrust comes practically entirely from the propeller, which has difficulties at such speeds.  Supersonic propellers have been tried (XF-88B and earlier), but they were found to be woefully inefficient because of shock waves.  Supersonic flow near the blades can occur well before the airplane itself reaches Mach 1.  That's the limiting factor.

I would add that speed of the propeller tips is the combination of the aircraft speed and the orbital speed of the tips, which is why the fastest turboprops use low RPM propellers.

I've read up that a turboprop draws in more air but at a slower speed and a turbofan draws in less air but at a higher speed.

While a turbofan ingests less air than a turboprop, both do it at essentially the same speed, which is flight speed. The difference is in the exit speed.
Thrust is air mass flow times speed difference between entry and exit speed. Engines of the same thrust will have the same product of mass flow times speed difference, but with the lower mass flow of a turbofan this only means that the speed difference is higher.
It should be immediately clear that a higher exit speed will also allow a higher flight speed. For thrust to be positive, exit speed must be quite a bit higher than flight speed, giving jets a larger margin than what turboprops can muster. In addition, the use of a propeller limits the useable top flight speed to clearly subsonic Mach numbers; a restriction that turbofans do not have.

The main operating limit for any blade is  reaching the transonic speed on its tip.
The tip of a blade sees an airflow's speed W which is the composition of the rotating speed U and the speed of flight V, as visible for example in the following plot from this report:

When this sum W reaches the transonic regime, there's a deterioration of the aerodynamic characteristics: lift decreases and drag and pitching moment increase, as visible for example in the following plots from this NASA report:

Note that this is valid for any blade: helicopter blades, propeller blades, fan blades of a turbofan and compressor blades of a turbofan/jet. They all are limited by the onset of the transonic speed on the blade tips (although the fans of some turbofans reach indeed supersonic speeds, but that's the result of different compromises).
Just as an example: the optimal rotation speed of an helicopter rotor is around Mach 0.7. This limits the speed of flight to what remains to reach the speed of sound i.e. the remaining Mach 0.3. That's why helicopters fly so "slowly". For a propeller the same reasoning applies, but the limitation on the flight speed is shifted toward Mach 0.6. And for the compressor's blades of a turbofan/jet that limit is somewhere in-between, around Mach 0.4.
Mach 0.4? But why then a turbofan/jet can fly supersonic, till Mach 3.2 for example for the Blackbird, in spite of that limitation? Because in front of it lies an extremely important component that most of the time passes unnoticed: the inlet. The main job of an inlet is to reduced the speed of the incoming air to Mach 0.4, whatever high the speed of flight is. If the speed of flight is high subsonic (Mach 0.4 to 1), then the inlet simply works as a divergent inlet (diffuser): its section gets bigger and bigger, the pressure gets higher and higher and the speed gets lower and lower (the compressor thanks Bernoulli). If the speed of flight is supersonic, in front of the inlet one or more shockwaves are formed, across which the speed reduces to high subsonic and afterward the divergent portion of the inlet finishes the work reducing again the airspeed till Mach 0.4. This is exactly how the inlets of a jetliner, the Concorde or the Blackbird work(ed) to supply their compressors with a relative slow airflow.


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Can an aircraft use ram air to generate power while flying?
I am a student carrying out a project focused on efficiency on aircraft. I have arrived at an idea I believe to be feasible. I am looking for suggestions, specification of what I would need, and any general tips and advice if this idea is actually reasonable.
I have found, using another page on this forum, the amount of fuel used by the engines on electrical generation. I am proposing to relieve this duty from the engines while in flight by opening large air intakes embedded into the rear of the fuselage on both left and right sides of the aircraft. These would guide air in and split into two convergent ducts that will spin a turbine that spins a generator. This power will be AC. It will be directly linked to the busbar system as a power source and feed into the battery after being converted to DC.
From an electrical and aerodynamic standpoint, is this feasible? Be as critical as possible. It is not a perfect idea, but the point is to save fuel by not generating power from the engines while in flight. This is aimed towards commercial airliners.

The approach is widely used in another transportation context - the (older) railroad rolling stock. The passenger cabin's lighting is powered by a generator rotated by the wheels. While it does not involve aerodynamics and is much more effective, it is still less effective than using wires between the carts. It is used because it reduces the complexity.

@MaxCremona its not clear from the question or your profile as to what level you're at in your studies. One thing I'd suggest, is to do a literature review, and find known inefficiencies and attempted solutions before getting attached to a given solution idea

This is a gentle reminder not to post (pseudo) answers in the comments. Please see this meta post.

@MaxCremona welcome to aviation.stackexchange.com

Just now found this highly related question -- https://aviation.stackexchange.com/questions/25933/is-it-sensible-to-harness-wind-energy-on-an-airplane -- also this one https://aviation.stackexchange.com/questions/32669/feasibility-of-using-secondary-props-to-generate-power-on-an-electric-model-airc?

opening large air intakes... These would guide air... that will spin a turbine that spins a generator.

You basically want to install a "ducted wind turbine" in the airplane.
Off the top of my head I see the following limitations/criticality:

Wind turbines are not as efficient as one might suspect: considering the aerodynamic and the electrical losses, an efficiency of 50% might be already a very good value. This value might be a bit higher since your turbine would be enclosed inside the fuselage i.e. it would be ducted. But, on the other side, since it would be located by the aft fuselage, it would work with highly turbolent airflow severely reducing its efficiency.

It would use up space and weight which could be used for the payload instead.

Something rotating is always imbalanced and therefore always producing vibrations; an even higher weight might be expected in order to cure potential resonance phenoma.

Something rotating can break and release debris possessing very high energy; the whole assembly would need to be enclosed in a solid casing to shield the structure; this obviously would add again weight.

It would work only when the aircraft is already moving with enough speed i.e. not on ground, nor at takeoff, nor at landing.

Everything on an airplane has to be regularly tested and inspected; the relevant cost would have to be added to the bill.

For sure there's something else that I'm forgetting.

Anyway there's a basic misunderstanding here:

I am proposing to relieve this duty from the engines while in flight... the point is to save fuel by not generating power from the engines while in flight.

Obviously the power extracted from the airflow by the turbine has been given to the airflow by the engines in the first place: so the wind turbine wouldn't work "for free" but it would anyhow consume power from the main engines possibly consuming even more power due to its drag and weight.
As also pointed out in other answers, something similar is actually already installed on aircrafts and is called ram air turbine. Anyway it is used only as a "last-chance" after a major power issue in order to get some electric or hydraulic energy to power the bare minimum systems to be able to land.

No, because large intake ducts would greatly increase drag.  The propulsion system moves the aircraft, so you are only indirectly and inefficiently drawing power from the engine/aircraft instead of using a simpler and lighter fan belt.
But all is not lost because internal combustion engines are hugely inefficient converting combustion energy into mechanical force.  Much heat is simply wasted in exhaust or radiated through cooling systems.
This waste heat energy could be recovered (as it is in steam boilers) by boiling expansion pressure --> turbine or piston motor --> condensation --> reboiling.
Recently a cascading boiling system with 2 or 3 lower temperature boilers recovering the heat from the previous condenser came to mind.
These would add weight and complexity, perhaps better suited for trains.
But the amount of heat loss is huge.  You may find success there.
Here is some information on power savings by removing an alternator from the engine.  Now one must see if additional weight of an energy recovery system adding to power requirements is greater or less than that saved from detaching the alternator belt.

The cascading boiler system sounds cool! Can you please provide a link?

@GeorgeMenoutis these were utilized on large ocean liners of the previous century.

@GeorgeMenoutis it is called recuperator and is normally used on micro-turbojets or on turbojets used to produce electricity (on ground). On aircraft turbojets it's not used due to the higher weight and complexity.

Engine efficiency, complexity, and weight are pretty closely linked. Weight directly impacts fuel consumption on an aircraft. More stages on the turbine means more efficient, but more cost/maintenance/weight. Bigger fan means more efficient, but more frontal area/drag/weight, and a slower turbine speed (unless you had a (heavy,expensive) gearbox).

Small antique aircraft which did not possess electrical systems originally do this already: on the bottom of the fuselage the owner can mount a little "windmill" propeller connected to a small generator. When flying, the windmill spins the generator and electrical power is carried into the fuselage with wires that are connected to a battery, which hence remains charged whenever the plane is in the air even though the engine does not possess a generator. Radios, lights, etc. are then connected to the battery.
Using a ram air turbine to spin a generator would be complicated and not very efficient. The right way is by gearing a generator right onto one of the plane's engines.

@user253751 The type of antique aircraft that neils is referring to didn't originally come with any kind of electrical generator at all. So the engine wouldn't have a convenient way to connect to an aftermarket generator, which means an external turbine is the only way to get things like lights, radio, transponder, etc. working without major engine modifications.

Today's airliners have a small emergency generator that is airflow driven.  It pops into action automatically when the last of the powerplants fails.  This electicity is needed to operate the hydraulics that operate the control surfaces, and the computers that control the hydraulics.  The plane is flown by wire and cannot be flown any other way.
The concern here is not efficiency.  The concern is the absolute need for power to fly the plane as a glider.  At best, this will result in a controlled descent ending in a very hard landing.  This is not good, but it's better than a crash.
There are a few cases where an airliner had to be glided to an airfield and landed without jet power.  These ram air generators worked.
Here are two accounts of airliners gliding to a landing.  One of them is at an airfield.
https://timeline.com/in-1983-two-pilots-miraculously-landed-a-jumbo-jet-with-no-fuel-from-40-000-feet-e51782deb01d
https://en.wikipedia.org/wiki/Air_Transat_Flight_236

I think this is an important piece of the answer: Yes, ram air turbines are totally doable and implemented on a regular basis but only as an emergency measure because they are totally inefficient (why add a lossy indirection if you can power the generator directly from an exiting engine!?). From a physics perspective one might add that they essentially convert "potential" (i.e., gravitational) energy of the airplane's mass at the given altitude into electric (or mechanical, via pressure) energy.

This is tangential to the question, but there's normally not any reason why an engine-out landing needs to be an unusually hard one. You can land more or less normally without engine power, aside from the obvious lack of ability to maintain altitude and airspeed at the same time and the inability to go around for another attempt. Of course, this assumes that you're in gliding range of a maintained runway, which the Gimli Glider and U.S. Airways 1549 were not. If you land somewhere other than at an airport, then, yeah, you'll probably have a hard landing.

It's an over-unity/"perpetual motion" myth
Something you regularly have to confront as a starting engineer.
Sort of like putting a windmill on an electric car to generate power to recharge the batteries so you arrive at your destination at 100%.
As they say, "there's no free lunch".  Your ram air turbine will increase drag on the aircraft, which will require more engine power to overcome.  So it is a net lose compared to a gear driven generator.
However, it is widely used when a module needs to be added to an aircraft that needs a lot of energy of one kind or another, and they don't want to enlarge the plane's electrical or hydraulic system to avoid a long re-certification process.

Intresting. Is there an example where this method was used to avoid re-certification process?

Finally a nice analogy with the car/windmill, I usually use a ship and a propeller. I've had many disappointing "inventors" at my desk when I explain them the physics of it. The answer to the OPs question is "yes, but not for free".

But actually a propeller on a car can add energy.  See the whole downwind-faster-than-the-wind paradox (google it, for example https://en.wikipedia.org/wiki/Blackbird_(wind-powered_vehicle).  Adding an electric motor and batteries would reduce efficiency hugely but in theory it would still work.  (Kidding, sort of.  It would only work when the wind was blowing.  But indefinitely sustained travel would theoretically be possible downwind as well as upwind.  The whole downwind-faster-than-the wind paradox is rather mind-bending to most of us... (hint: it can't work in an airplane though!)

@quietflyer I found that whole "Woo woo, wind powered car exceeds speed of wind" thing to be a case study in the importance of viewing a problem from the proper perspective/POV.  From the car's frame it is obvious lol.  But good point, ambient wind breaks my metaphor.

@AdityaSharma ECM jamming pods on combat aircraft.

Actually the car that goes faster into the wind proves lift can be greater than drag for airfoils.  But loss of efficiency converting wind into electricity makes it a non-starter.

@RobertDiGiovanni Yes, that confuses me, since torque->electric->torque transmissions are very efficient. Look at any diesel locomotive - the cooling in main generator and traction motors is quite modest. Since electric is not significantly less efficient than gear, belt or whatever the Blackbird used... why not? The only thing I can think is if you do put a windmill on an EV, every police officer will stop you and cite you for violating the First Law of Thermodynamics :)

@Harper-ReinstateMonica well, with cars, acceleration is more the name of the game.  0-60, pick-up.  But ice boats seem to have the faster than the wind thing down really well.  And our turbine car only slowly accelerates.  I have heard alternators generate a lot of heat, but an old-style generator ... the opposite of an electric motor ... perhaps.

@RobertDiGiovanni I was thinking of a windmill on an EV.  I mean anyway for harvesting when it's parked. But the Blackbird principle says it could work while driving too.  The blades need to be optimized for 60-120 mph instead of 0-40 mph though.  I'm sure car alternators are not efficient, they're all about price point.  Locomotive alternators spare no expense because the customer is all about fuel savings and can capitalize the alternator for 27.5 years.

@AdityaSharma Also, fuel pumps on refueling pods. Carriers are very cramped places and the US Navy was on a kick to reduce the number of types of airframe a carrier must support. So they scrapped all their air-tankers and created a kit that bolts onto the F-18 Hornet to turn it into a tanker. That uses a RAT for a fuel pump. And their plan didn't work anyway, since they are wearing out the F-18 fleet...  and still have to support the C-2/E-2 and they would've made fine airtankers.

@Harper-ReinstateMonica Intresting information, thank you.

Yes. Early F-4 aircraft had Ram Air Turbines, (we called them RATs), that provided emergency electrical and hydraulic power in the event of system failures. They were small propellor driven hydraulic pumps and electrical generators/(or alternators, not sure)
The RAT would deploy, when you actuated it, into the slipstream from a compartment in the lower fuselage

+1 for mentioning existing RATs, but you might want to add that these do not save fuel. Also, since the question was about commercial airliners, RATs are quite common on airliners as well (Boeing: since 757, Airbus: all).

Bianfable.. not sure what you are saying (about saving fuel). In the F-4, they were there to provide hydraulic (& electric) power in emergency situations.

@CharlesBretana Consider reading the question :-)

The only situation where this could make sense is when the plane descends from travel altitude as some kind of recuperative "braking". But the weight of the extra gear needed would make this viable only for very short flights, on most the time spent in descent is simply too short to have any impact on overall energy usage. Additionally, it wouldn't be enough to eliminate the need for thrust reversal.
This could, however, get interesting for electrically powered planes where the engines' power flow could simply be reversed, needing no extra parts, while also being able to store the energy in the - at that point - mostly depleted batteries, making every bit of generation a net gain.

Good point about electrically powered airplanes. Anyway I'm not that sure that the same circuitry can be used both ways

@sophit: The power electronics would have to be designed for it, but that's relatively light-weight.  I think most electric motor designs can Just Work as generators, converting kinetic energy to electrical instead of the reverse.  It's well worth it in cars where stop-and-go is common, and any extra weight that is required doesn't cost cruise efficiency, except a tiny bit of extra rolling resistance from the tires. Neither of those things are true in airplanes.  So it would only be worth it if the extra weight is minimal. (And cost of engineering design, maintenance, and reliability).

@PeterCordes: thanks for your interesting answer. I know that in real-size electric aircrafts inverters+relevant electronics need massive heat exchangers. Can those inverters be used both ways or do we need another set?

@sophit: You don't need a (true RMS) inverter, that produces AC voltage.  To charge a battery, you just need to rectify and filter to DC.  Also keep in mind the duty-cycle involved: occasional bursts of energy recovery in cars give plenty of time for things to cool down.  (A temp sensor could disable energy recovery if things are getting too hot, but I expect it's fairly efficient so not a lot of energy lost as heat).  In a plane you might be doing energy-recovery for a longer descent, but probably at a lower power level.  And air cooling is easily available; it can only operate while moving.

@PeterCordes

Yes.  Ram Air Turbines (RATs) are very common in aviation for applications where equipment needs electrical or hydraulic power that aircraft engines cannot provide.
The most common application of a RAT is an emergency module, usually stored retracted and faired, but can be extended into the airstream.  They are fitted with a RAT which drives an electrical generator and/or hydraulic pump to provide emergency electrical or hydraulic power for actuation of primary flight controls in the event of a major systems failure.

Another application for RATs is to deliver electrical power to mission system modules independent of ship’s power.  A common specific application of this is the ALQ-99/E jammer pods carried by both the EA-6B and EA-18G aircraft.

You can see the RATs in operation on the EA-18G aircraft below.

The RAT on the nose of the pod drives a generator providing electrical power to the pod’s systems. Advantages to the system are the ability to carry and use these pods without a beefed up ships electrical system and additional power distribution wiring in the airframe as well as flexibility in mission configuration.  The downsides are drag, generally resulting in a 2-3% range reduction per RAT in operation.  And it’s isn’t a very stealthy option, either.

Here's a concept inspired by this related answer -- imagine an airplane that is essentially a sailplane with an electric motor for added assistance.  Normally, once a sailplane has climbed to the top of a thermal updraft, there is no way to extract any more energy from that updraft.  However, our new concept motorglider will use the propeller and motor as a generator to charge a battery while climbing, accepting a lower climb rate for the sake of extracting more energy from the thermal updraft.  The same can be done in wave lift, etc.  Wave lift in particular would be conducive to simply remaining at a fixed altitude where the updraft is strongest, and extracting energy with the generator at a very high rate, until the batteries are fully charged.
The purpose of our new concept aircraft will be to make very long-range flights-- potentially trans-continental or trans-global, and potentially spanning multiple days-- using no energy other than atmospheric energy.  This could be done for the sake of competition (flying against other similar aircraft), or to attract publicity to various causes, or simply to advance the state of aeronautical technology.  A new category of FAI record might be appropriate.
In order to meet the stated goal, take-off would be conducted with the batteries containing only enough charge to climb to a limited altitude representative of a normal aerotow, or else the batteries would be drained after the initial climb before proceeding with the rest of the flight.
A logical extension of the concept would be to also include solar panels, in which case the goal would be to fly using only external, natural energy harvested during the course of the flight, be it "direct" solar, or "indirect" solar (i.e. atmospheric).  Or if the "using... only energy harvested during the course of the flight" portion of the concept were dropped, the batteries could be fully charged (via solar energy) before takeoff and freely used as needed before in-flight recharging by both available methods.  Another variation of the concept would omit the solar panels, but would fully charge the batteries before takeoff using energy harvested from wind turbines.
This thought experiment should make it clear that the answer to the original question--

Can an aircraft use ram air to generate power while flying?

as well as the implied question "would it ever make any sense to do so?"
Is "yes".

It could work, but you need to carefully estimate the impact of increased drag. In majority of motorgliders, the whole engine block or at least propeller blades have to be folded when not in use - otherwise they hardly can climb.

As some of the commentators stated - theoretical answer is "yes", but practical is "it will be less effective".
If we compare two systems - main engine(s) pulling the air through and causing this air to rotate the turbine vs main engine(s) rotating generator(s) via mechanical gears - the losses in the former one will be much higher, due to "openness" of the system, turbulence etc.

Welcome to aviation.SE. I fail to see how your answer is adding anything that was not said in the other answers already. Would you mind expanding it?

Some citations etc needed here - there is plenty of reason to think that a well-designed RAT could be slightly more efficient than an engine-driven generator, since it eliminates mechanical losses.

@MikeB it eliminates minor mechanical losses and introduces major aerodynamic ones. You can do a simple test - put electrical motor with propeller against the same-size propeller that can spin freely. Then measure the rotation speed of both - the difference will tell you how much energy is lost in transmission of the power.

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Why does Engine Pressure Ratio (EPR) rise as thrust levers are pushed forward?
I'm looking for an answer that at least addresses the fundamental thermodynamics of a jet engine.
Jet engines use a Brayton cycle to get useful work out of a heat source. An ideal Brayton cycle engine has the input and output pressure be the same. If Exhaust pressure is greater than inlet pressure, then that's becoming closer to Diesel cycle further from a Brayton cycle (And to those who feel like correcting this point, a diesel cycle has many forms beyond what is typically thought of as a diesel engine but that's not want I want to talk about. Diesel cycle only cares about adiabatic compressions/expansion, constant pressure combustion, and V1 = V4, but I do not wish to dwell on this).
Also, when looking at nozzle design, we ideally try for the exhaust pressure to be equal to ambient pressure as opposed to an over expanded nozzle where the exhaust pressure is less than ambient or an under expanded nozzle where the exhaust pressure is more than ambient.
Both of these two things together seem to indicate that a jet engine would be designed to have the intake and exhaust pressure be the same (ie the EPR should be 1). Is there something that would make achieving this in the real world challenging? Or are the engine designers not even trying to have the exhaust pressure be equal to the ambient pressure? If the latter is the case, why would you not want to try to have the engine operate as closely to the ideal cycle and nozzle design as you could? If it's the former, what conditions would contribute to this becoming more challenging?
I get that if an engine has fixed geometry, if you change the amount of heat added in combustion, it will change the EPR, but there are plenty of jet engines have variable exit geometry. Therefore, I would think you could vary the geometry to get the ideal exhaust pressure. Even if you couldn't vary the engine geometry and only design for 1 power setting, I would expect that it is most efficient where the engine is expected to be used the most. For example, if I expect my engine to spend most of its time being used at 35,000 ft and generating a specific amount of thrust, I would design the nozzle so that the discharge pressure would be whatever ambient is in those conditions. That would translate in the real world to the engine operating at or very close to EPR of 1 during cruise.
It is my understanding that this is not the case, and if that's a correct understanding, I am missing a piece to the puzzle to make sense of why things are the way that they are.

Related reading here.

@RobertDiGiovanni Not quite related, this isn't about FPR or BPR but EPR... why all those abbreviations! ;-) FPR and BPR are design parameters, EPR is a control parameter.

What makes you think the EPR should be 1.0? Note that we are using stagnation/total pressures to determine the EPR, we are not using the static pressures. Although the nozzle expands to ambient pressure, the total pressure is significantly higher due to the fact that the nozzle is producing a high velocity jet stream. The velocities are quite different from the inlet and exhaust mass flow! If not you do not have thrust. EPR is a measure for how hard the engine is working and from that we can derive how much thrust is generated (or if the thrust level is always the same, regardless of engine condition). The higher the EPR the harder the engine works (more fuel is added the faster the spools spin). In afterburning turbofans the use of EPR can be used in afterburning mode to schedule the nozzle area (EPR is kept constant during AB mode); this basically maximizes the spool speeds of the turbofan to MIL power (if you would not open the nozzle, the fan would stall).
Do note that the EPR is defined as the nozzle exit pressure over the fan intake pressure, but more commonly the mixer (where fan and core flow meet in the afterburner) or HPT turbine exit pressure (both total pressures) is used for calculating EPR, these are more easily measured.
We can experiment with the ideal nozzle expansion using an afterburning turbofan model in a (dry/non-afterburning) simulation.

The latter image shows the representation of Pt6 over Pt2 in the upper part and the Pt9 over Pt2 in the lower part.
The following image shows the EPR values for a deteriorated engine as function of thrust, you see that EPR is a direct measure for the thrust level. A deteriorated engine has less thrust for the same combustor exit temperature, and also lower EPR value. This makes the EPR parameter valuable for engine control purposes.

Keeping EPR constant (at the MIL power setting) and increasing the fuel flow in the afterburner and opening the nozzle (calculated area based on added fuel and constant EPR) you will see the thrust increase the more fuel is added:

I did not realize that it was stagnation pressure not static pressure, but then adding afterburner would raise the EPR, which it sounds like you're saying it doesn't. The whole point of AB is to raise the heat of the exhaust, which means we can extract more kinetic energy, which means more thrust. But that higher speed of the exhaust will have a higher stagnation pressure than it otherwise would, so the EPR should rise when AB is on when compared to mil power should it not?

@youguysfail No, it would not rise, it is kept constant by opening the nozzle. In fact the whole engine in front of the AB doesn't even realize the AB is on, the engine runs in MIL power setting. If you wouldn't open the nozzle during AB, yes the EPR would rise, but then the fan will stall. In AB the EPR is the same as in MIL power, for every AB setting.

Now I'm even more confused. If the pressure is the same in the exhaust both with and without the AB, then the stagnation pressure with AB MUST be higher with AB than without. This is because stagnation pressure is static pressure + .5densityvelocity^2 [Or it's P(1+(k-1)/2M^2)^((k-1)/k) for compressible flow, either way static pressure goes up as velocity goes up]. Since AB raises the velocity but keeps the static pressure the same, that means the stagnation pressure has to go up too. If the stagnation pressure goes up, so too does the EPR. Where am I going wrong here?

The EPR defined by Pt6/Pt2 is kept constant, as described, how would you measure the pressure at station 9, hence the used EPR metric. Static pressure goes down due to afterburning, see e.g. https://aviation.stackexchange.com/a/91876, due to heating you introduce an additional pressure loss. I think EPR (if defined as Pt9/Pt2) goes down instead of up, download a simulation program and see for yourself.

The simulation is useful but I'd also like to understand they key general key concepts here. Thanks for your help though it is much appreciated.

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Calculate propeller efficiency
I have a test bench that can test the thrust of a propeller with my electric motor, is there a way that I can calculate the propeller efficiency (ηp)?

Only if you can simultaneously measure how much power the motor draws, and you don't mind correcting for the many other factors mentioned elsewhere on this site.  If you're measuring static thrust (zero airspeed) and don't have a wind tunnel, there are ways to estimate in-flight efficiency at various airspeeds, too.


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Why does the gas turbine recuperator hugely increase efficiency but slightly decrease the peak power?
I understand that the heated intake air will burn more efficiently with the fuel and that's why recuperators increase efficiency. The part I don't understand is that if efficiency is going up, why isn't the power output also not increasing along with efficiency? For example, if we used to make 500 hp from a free shaft gas turbine by using 100x amount of fuel and then we add recuperator and again add 100x amount of fuel, wouldn't the engine now make much more power than the original 500 hp with the now more efficient combustion thanks to the recuperator?
Is there a limitation in oxygen rich air supply like in piston engines so that we cant add as much fuel when there is a recuperator added? Or is this caused by some reason related to the turbine itself and its ability to extract energy from the exhaust flow?

Hot air is less dense, so the engine has a lower mass flow when using recuperation. It would be great to know more about the specific setup you have in mind; this would allow me to actually give a precise answer.

I understand that the heated intake air will burn more efficiently with the fuel and that's why recuperators increase efficiency.

Actually that is not the reason.
When applying a recuperator to a gasturbine design, the heat from the exhaust is used to warm up the combustor entry air, the effect is that in order to reach the same combustor exit temperature you simply need less fuel, that is the main reason why recuperated cycles are more efficient.

The part I don't understand is that if efficiency is going up, why isn't the power output also not increasing along with efficiency?

Why would power go up, it is expected to go down actually! The reason for this is that you cannot add more fuel to the existing design as that will cause the exit combustor temperature to increase, the turbine cannot handle those temperatures. So, with fixed turbine entry temperature you cannot add more fuel. What you do with recuperation is add more complexity to the cycle (heat exchanger, tubing, etc.).

Is there a limitation in oxygen rich air supply?

No, there is plenty of oxygen available, but you cannot use it because the temperature of the turbine would get too high.

Why does the gas turbine recuperator hugely increase efficiency but slightly decrease the peak power?

The first part of this question is already answered above, the reason for a decrease in power is related to the insertion of the recuperator. A recuperator is a heat exchanger and has pressure loss deltas over the 2 sides of the heat exchanger, pressure loss cause the output power to decrease.
This is best shown by use of a simulation example. Assume we have a turboshaft gas turbine without a recuperator and assess the performance. This is best done by sweeping the combustor exit temperature, this is then in turn used on a recuperated engine with and without pressure losses in the recuperator.
Below is a graph of such a simulation exercise in a gas turbine simulation tool. The three graphs from top to bottom show the specific fuel consumption SFCshaft, combustor exit temperature Tt4 and the combustor fuel flow Wf_b plotted against shaft power PWshaft. The red dot-dashed line shows the performance of a non-recuperated turboshaft engine. The blue solid line is an ideal recuperator operating without pressure losses and the black dashed line account for a pressure loss in the 2 recuperator flows.

From the graphs you see that the specific fuel consumption decreases when using a recuperator, this is clearly made visible in the fuel flow graph where the recuperated cycle requires less fuel. Furthermore, the pressure loss in the recuperator shifts the curve to the left, so less power is generated for the same turbine entry temperature sweep.

I had a wrong approach at first as you have mentioned, i understand it very clearly now, Thank you so much for the answer


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Why do the lancet drones for russia have x-wing?
I was looking at some videos of the war. The russian ZALA Lancet drones have an X wing configuration.
Why?
I looked up this book. It has a chapter on the X wings. Which says, that X wings can generate double the lift than single wing, but also upper wing barely generate any lift.
Is there any document explaining the wings and design decision of this drone?

Can you tell us in which chapter/article of the book you linked you have found that claim? The book is quite big.

@sophit
Page1176: The Aerodynamics and Hover Control for the [sic.] X-wing VTOL MAV

Having two wings staggered above one another is always disadvantageous from an aerodynamics point of view, as the two wings do not create twice the lift as an equivalent mono-wing (see biplanes). Additionally, having a second wind creates a second wing-tip vortice which increases drag.
However, the lancet drone is not a conventional aircraft but a loitering munition weapon, designed to be able to hit a ground target with high accuracy, which is why the designers opted for an X-wing design. The big advantage of this design, is that the lancet drone does not have to bank-to-turn, but can employ a skid-to-turn strategy (see also this book). This allows the lancet drone to adjust its movement direction faster than a conventional (mono-)plane. This means that during its final approach phase, the lancet drone can make course corrections not only by having to roll into the desired direction and then "pulling on the stick" but rather it can just "sideslip" - which decreases steering response time resulting in a more accurate guidance and therefore increases accuracy of the weapon.
To sum up, this design decision trades aerodynamic/power efficiency for accuracy.
So why is an X-wing type design well suited for this skid-to-turn strategy? Because with X-wings, you can generate lift not only in the vertical direction, but also laterally. More or less, X-wing designs are capable of changing flight directions regardless of their orientation, which enables them to fly up, down, left or right without having to roll in these directions first.
This strategy is also commonly applied by conventional missiles, for the same reasons. Additionally, it makes the GNC system simpler, as you simply deflect your fins into the direction you want to go!
Of course, I cannot produce a link or a document to proof this, as all of the design documents of this weapon are classified.

Hi @U_flow, thank you. But can you point me some work explaining X-wing skid to turn? Googling the term "Xwing slip to turn" is not returning anything but some movie CGI stuff.  Thank you.

@Sean I posted the wikipedia article explaining the Skid-to-Turn strategy in the answer (https://en.wikipedia.org/wiki/Skid-to-turn). Basically it just means that instead of first rolling to initiate a turn, you directly point your aircraft in the direction you want to go. Note that this is different from how a normal aircraft flies mostly. I tried to update my answer a following your comment above. However I am uncertain what you are confused about, therefore please ask and I will try to clarify my answer further.

For further explanaition please also see this book.

thank you. I just want to understand the concepts more deeply. my background is computational fluid dynamics. But I am interested in understanding the design choices behind these typeof things.

I'm sure you wanted to say "which decreases steering response time".

Yes, that is true. I will edit accordingly.

"Additionally, having a second wind creates a second wing-tip vortice which increases drag" having a second wing-tip vortex doesn't imply higher drag. Actually if the lift is shared among two wings like a biplane then the induced drag halves, at least theoretically.

@sophit to my understanding, the formation of a second wing tip vortice creates higher drag in comparison to a single one, as there is "overhead" included. But I leave the exact explanation to someone with more aerodynamic knowledge then me.

Simple mathematical explanation: induced drag is proportional to $C_l^2$. If you have two wings like in a biplane then each wing is generating half of the total lift and the relevant induced drag become $2 \times (\frac{C_l}{2})^2=\frac{C_l^2}{2}$ i.e. half of before

@PeterKämpf can you share your insights on this topic?

@PeterKämpf: please charge for your service

@U_flow At high speeds I do not pay much attention to induced drag, because friction is dominant. Having two pairs of wings where one would suffice will increase drag. Induced drag is important in the loitering phase, and here I expect not much difference to a single wing layout - if you look at the mass of air which is used for lift creation, it always is a circle with the wingspan as its diameter.

Why do the lancet drones for russia have x-wing?

They not only have x-wings but a canard configuration and a pusher-propeller as well. And everything points to some very good aerodynamics choices. Anyway wings and tails are not really arranged in a "X" rather in a configuration that I would call "double V" (or "><"): in fact if you rotate the aircraft 45° you won't get a perfect cross + since the angle between the two wings is not perpendicular.
The concept of using two wings one (more or less) on top of the other and the concept of the canard configuration both are aimed at reducing induced drag. Since the Lancet is flying at a relatively low speed, induced drag is going to be the highest source of drag during flight and it makes sense to reduce it as much as possible to extend its range and speed.
Theoretically, induced drag is proportional to the square of the lift $C_{d_i} \propto C_l^2$. Now, if you split the total lift among two wings, each carrying half of the total lift, then the relevant induced drag becomes:
$C_{d_i} \propto 2 \times  (C_l/2)^2 =\frac{C_l^2}{2}$
i.e half of before! And since the lancet actually has four lifting surfaces, the reduction in the induced drag is going to be of a factor $\frac{1}{4}$.
This result is obviously only theoretical: aerodynamic interference among the four lifting surfaces and the need for longitudinal stability will increase parasite drag and will prevent the perfectly equal sharing of lift among the four surfaces; anyway there should be a net gain in induced drag compared to a conventional configuration.
The pusher-propeller located on the tail is also helping in stabilising the aircraft longitudinally, further downloading the canard.
The fact that all four surfaces are (or seem to be) movable will also give an extremely manoeuvrable aircraft.


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What does fan pressure ratio affect in an engine?
So I'm doing a project for my engineering project and nowhere can I find what the pressure ratio actually affects, in this digital sim I am using, when I adjust the FPR it has zero effect on any outputs so what does it actually do?
Here is the link to the sim if anyone wants to check it out.

I suspect you're leaving BPR at 0.0.  That means no bypass and no fan -- so FPR won't do anything.
Increase BPR and you'll find that FPR is now 'active'.

I had some good luck with TSFC using a bypass ratio of 10 and FPR of 1.4 at Mach 0.85.  Apparently, there is some link between FPR and core pressure as FPR ratio > 1.6 choked the engine airflow too much.  Lower FPR than 1.2 produced more thrust but higher TSFC.

@RobertDiGiovanni It totally depends on the complete engine cycle what the effect is of changing the FPR is. Parameter sweep studies are an excellent choice to see what happens rather than changing only FPR, see my answer below.

It seems you never looked at efficiency, or TSFC (thrust-specific fuel consumption) in that simulation.

The Fan Pressure Ratio (FPR) is part of the Overall Pressure Ratio (OPR, ratio of highest to lowest pressure in your primary gas path from inlet to combustor).
The importance of the FPR depends on the application of your gas turbine, generally, higher pressure ratios are found in smaller fan diameters with lower flow capacity while lower pressure ratios are found in larger diameters with high flow capacity. Why? The larger the amount of air is accelerated at a smaller velocity increase the higher the propulsive efficiency. This is where the ByPass Ratio (BPR) comes in, this is defined as the ratio between the mass flow rate of the bypass stream to the mass flow rate entering the core.
An excellent study to visualize the effect of FPR and BPR is done by Dr. N. Cumpsty in Preparing for the Future: Reducing Gas Turbine Environmental Impact. In revisiting this study for more recent engines, a gas turbine simulation tool can be used to show the effects of changing the FPR for a series of values of BPR. This can be an excellent starting point for the design of new or derivative gas turbines to find the optimum (where the Thrust Specific Fuel Consumption, or TSFC is lowest).
E.g. for a more recent (study from 2019) Ultra High Bypass Ratio turbofan engines you will get TSFC as function of BPR for FPR values from 1.6 down to 1.3 in steps of 0.05:

Basically, the FPR and the BPR determine where the optimum TSFC is for a particular engine (for a fixed technology level of your engine, e.g. choice of thrust at Top Of Climb, Tt4, Turbine cooling mass flow and OPR). If BPR is fixed, lowering FPR either increases or decreases the TSFC, it depends on your choice of BPR.

Please note that changing the FPR (or BPR) implies a change of the engine configuration (also known as the design), the tool referenced in the question only specifies TET (Turbine Entry Temperature, also known as Tt4) through specifying the FAR (Fuel/Air Ratio), preferably you want to constrain the thrust value as you need to design for a specific application/performance point. The webtool is limited and should be used wisely. Exhaust area is a calculated parameter in gas turbine design calculation/simulation, in the webtool it is a variable/scaling parameter (usually thrust is used, which is far more intuitive/practical than the exhaust area). In recreating a J85 turbojet engine it is confirmed that the webtool doesn't predict the correct performance, 20 - 40 % more is seen depending on the parameter.

Since it is the only dimensional parameter, it seems exhaust area is simply scaling the engine. As you say, the webtool is limited and should be used wisely. +1


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Is there an aerodynamic force that would keep this experimental WW2 era prop from flying as fast as an early jet?
The XB-42 was an experimental bomber developed during WW2.  Its twin engines delivered a combined 3,800 HP and top speed of 410 mph at 23,440 ft.

It's an amazing aircraft.  But if an enormous aircraft is that fast why not use the same concept for smaller fighter-sized aircraft?  I assume there must be a good reason because it didn't happen.
I used a modified version of the drag equation found on NASA's website to estimate the speed increase if the front profile was reduced by 50% and everything else stayed identically the same.  The modified version can be found here.
By reducing the front profile by 50% the top speed increased by 25.99%.  That brought its top speed up from 410mph to an amazing 517mph.  That's almost as fast as a jet.
Is that possible or am I missing some other aerodynamic force that would stop the plane from approaching that speed?
I guess at the most basic level is this concept correct?
Smaller frontal area = faster top speed.
Is there some other aerodynamic force that would negate that?
Answer to comment below: Zeus answered that reducing the aircraft's profile wouldn't make a significant difference because most drag was produced by the wing.  Yes and no.  Wing lift does produce a lot of drag at low speed but that drops as the plane accelerates.  At high speed nearly all drag is parasitic (i.e. the shape of the aircraft) as evidenced by the chart below.  Moreover my question was theoretical.  If 50% of the plane's profile was removed the weight could be removed as well and with it the wing size.  Turning the XB-42 airframe into a small, fighter aircraft was the general point of the question.

It looks amazing but I think we have the answer

For starters, unlike with cars, most of the drag is created by the wing, and much of it is the by-product of lift. And since you you didn't mention you were going to reduce weight, you can't reduce the wing...

@Zeus please post an answer

This question is really unclear because the speed of "a" jet could be anything. There are fast jets and slow jets, and in fact quite a few of them fly slower than 500mph.

@DR01D The speed of sound is not a constant, but a function of temperature and the thermodynamic properties of the gas (air in this case). The speed of sound decreases noticeably as altitude increases.

@DR01D Not disputing that, it seems accurate, and I was mostly nitpicking your previous comments. However, transonic effects start to be relevant below Mach 1, for example, modern airliners fly at around $M = 0.85$ and require careful wing design to avoid excessive drag due to local supersonic areas on the wing.

First of all, there were several propeller aircraft which were much faster than the XB-42. But let's stick with it and scale it to see where that brings us.
For that we need two things: Equations for drag $D$:
$$D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\left(c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right) = \frac{\rho\cdot v^2}{2}\cdot S\cdot c_{D0}+\frac{(m\cdot g)^2}{\pi\cdot AR\cdot\epsilon\cdot\frac{\rho\cdot v^2}{2}\cdot S}$$
and for thrust $T$:
$$T = \frac{P}{v}\cdot\eta_p$$
The symbols are (values are for the XB-42 at 7,140 m):
$\kern{5mm} \rho\:\:\:\:\:$ air density = 0.581 kg/m³
$\kern{5mm} v\:\:\:\:\:$ velocity = 410 mph = 183.3 m/s
$\kern{5mm} S\:\:\:\:\:$ wing surface area = 51.6 m²
$\kern{5mm} c_{D0} \:$  zero-lift drag coefficient
$\kern{5mm} c_L \:\:\:$  lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$  3.14159$\dots$
$\kern{5mm} AR \:\:$  aspect ratio of the wing = 8.95
$\kern{5mm} \epsilon \:\:\:\:\:\:$  the wing's Oswald factor = 0.9
$\kern{5mm} m \:\:\:\;$  aircraft mass = 15,000 kg
$\kern{5mm} g \:\:\:\:\;$  gravitational acceleration
$\kern{5mm} P \:\:\:\:\:$  engine shaft power = 3800 hp = 2833.66 kW
$\kern{5mm} \eta_P \:\:\;$  propeller efficiency = 0.85
In order to determine the unknown zero-lift drag, drag is set equal to thrust, the values above are inserted and the equation is solved for $c_{D0}$:
$$c_{D0} = \frac{\frac{P}{v}\cdot\eta_p - \frac{(m\cdot g)^2}{\pi\cdot AR\cdot\epsilon\cdot\frac{\rho\cdot v^2}{2}\cdot S}}{\frac{\rho\cdot v^2}{2}\cdot S} = 0.02272$$
which looks about right.
Then we need some clever formula how mass will change with a scaled airframe. Since both the engine and the pilot will stay constant, to assume that mass will scale with the square of the length is quite generous. So we have a second aircraft of 70.7% size which will have half the frontal area, half the mass but the same power. Now we need to solve for the unknown speed with the now known zero-lift drag:
$$v^4 = \frac{2\cdot P\cdot v\cdot\eta_p}{\rho\cdot ½\cdot S\cdot c_{D0}} - \frac{(2\cdot ½\cdot m\cdot g)^2}{\pi\cdot AR\cdot\epsilon\cdot\left(\rho\cdot ½\cdot S\right)^2\cdot c_{D0}}$$
$$v^4 = 14.144.659.448\cdot v - 167,497,545.8$$
$$v = 237.76\:\text{m/s} = 531.85\:\text{mph}$$
Now all what is left to be solved is how to squeeze the existing engines and their cooling system into half the fuselage cross section.

I guess at the most basic level is this concept correct?

To answer your question directly: I used a slightly different approach and got an even higher speed. If I leave mass constant, the higher induced drag will limit the speed to 223.37 m/s, which is just shy of 500 mph. I wonder how you got a speed of 517 mph. The car example on the linked site does not include induced drag, so I guess your concept needs a little improvement.

What would you say about the prop(s) required to drive such a hypothetical aircraft?  Rough calc's say the XB-42's props would have to support a pitch of at least 237.8m (i.e., blades nearly edge-on to the direction of flight) just to generate any thrust at all at 531.85mph.  Feel free to correct my math. :)

@Peter Kämpf Perfect thank so you much!  That was the equation I was looking for.  In this case I think that yes the engines and cooling system would fit in a smaller aircraft.  The XB-42 had a huge bomb bay that held 8000 pounds of bombs and a crew of 3.  There is plenty of fat that could be removed from that plane to bring it down to fighter size and still leave room for both engines.

Thanks for the link to the XF-12 Rainbow!  https://en.wikipedia.org/wiki/Republic_XF-12_Rainbow I hadn't heard of that aircraft before.  Really amazing bomber.

@DR01D: The XF-12 was a reconnaissance aircraft. Not every four-engined airplane is automatically a bomber!

@ioctlLR: Yes, you would need more prop solidity to absorb the power and a lower prop speed in order to avoid supersonic tips. The efficiency will most likely drop - after all, this is already Mach 0.767. But 520 mph should be feasible with some adjustments.

@Peter Kämpf  My concept was just a thought experiment.  I didn't include induced drag because I wasn't sure how to.  Thank you so much for your equation!  I'll use those in the future.

@Peter Kämpf I'm using your thrust equation on the XB-42 and could you tell me if I accounted for this correctly?  T = (2833.66 * .85)/183.3.  And this gives me 13.14 units of Thrust.  If I did that correct do you mind if I ask what are the units?  Thank you so much!

@DR01D: If you use kW for the power, the result is in kN force. 13.14 kN at 15 tons mass gives you a thrust/weight ratio of 0.089, so the L/D at max speed is 11.2. This looks about right.

@Peter Kämpf  Wow! I didn't realize that early jets only produced about 1/3rd more thrust than a prop.  https://en.wikipedia.org/wiki/Junkers_Jumo_004  The Jumo-004B that powered the ME-262 produced 8.83Kn, 004C produced 9.81Kn and the 004D produced 10.3Kn.  That's really interesting.

Let us continue this discussion in chat.

@Peter Kämpf  Hi Peter, if I used your amazing answer found here, https://aviation.stackexchange.com/questions/420/how-can-i-calculate-maximum-rate-of-climb to determine the XB-42's climb rate would it be possible for you to check my result just to make sure I didn't screw something up?  I can put the answer in that thread.

@Peter Kämpf Hi Peter, your equation is amazing but I think you might have made a tiny mathematical error at the end.  x^4 = 14144659.448 * 183.3 - 167497545.8 = 221.9155 not 237.76.  Maybe I did something wrong?  Could you confirm either way? Thanks so much, this is really helpful!

@DR01D The second $v$ is the same as the $v$ in $v^4$ on the left side (and not 183.3). I simply put the equation into Numbers and adjusted $v$ until the result was equal on both sides.

@Peter Kämpf Oh.... I get it.  v^4 = 14144659.448 * v - 167497545.8.  Try different Vs until it works on both sides.  Awesome, Thanks!

+1. Insightful answer!

all propeller-driven aircraft are subject to the same velocity limit. to make the plane go fast, the propeller has to spin faster, and the propeller blades achieve mach 1 before the airframe does. as they go supersonic, their thrust output goes down and they waste work in making noise instead of propulsion. the propulsive efficiency loss is sufficient to prohibit the airframe from achieving mach 1.

those forces are sufficient to stop a prop driven aircraft from reaching mach 1.  But would they in theory be sufficient to stop a plane traveling at 410mph from reaching 517mph?

well, have a look at the relationship between top speed and horsepower first. HP requirement goes up as the cube of airspeed. can you fit that HP into the existing airframe?

In this case I think yes.  The bottom portion of the XB-42 was a big, empty bomb bay that held up to 8,000 pounds.  With that gone you could reduce the wing area and shorten the tail.  Plus a WW2 fighter only needs 1 pilot and the XB-42 had a crew of 3.  So with all of that gone you could maybe reduce the size by half and still keep the same engines and HP.

You can increase the pitch of the prop and by doing so you can keep its RPM low (subsonic). So the propeller tip speed problem can be a trade off and can be potentially solved.

@gurkancetin, i think not, for the following reasons. on an airplane that is moving through the air at, say, 1000fps, the relative speed of the rotating propeller moving through that air will always be greater than the speed of the plane. it is inevitable that the prop blades will hit sonic velocity before the airframe will.

The DO-335 was a ww2 era prop that flew 475mph.  How did they solve the prop problem?  https://en.wikipedia.org/wiki/Dornier_Do_335

One factor to remember is that power input dictates the total size and blade-count of the propeller.  The XB-42 had two 13ft three-bladed props, which means the propellers' red-line RPM (mach .92) would be roughly 1500rpm, sitting still on the ground.  You'd have to drop that RPM quite a bit at 410mph and FL240 to stay under the red-line.

@DR01D Voodoo posted 531.53 mph average in four laps last year at Reno, with a fastest pass of 554.69 mph.  That's with a really big, really high-performance prop, a "metric ton" of horsepower, and a highly-optimized airframe.  It's possible to go that fast, but not practical; Voodoo is effectively purpose-built to go fast and nothing else.

@ioctlLR That is interesting, I wasn't aware of the prop limitation.  But the DO-335 flew at 475mph back in 1944.  That's almost 75 years ago.
So clearly there must be a way to solve that problem.

@DR01D The airflow separates from the prop blade at mach .92, causing the blade to vibrate (destructively, if pushed any further) and a bunch of energy to convert to sound instead of thrust.  No propeller design can get past that limit (hence the reason everyone called it the sound "barrier").  Even at mach .91 the prop efficiency is so bad it's not really accelerating the aircraft any more.

@ioctLR No doubt you are right but I don't mean modern jet speeds.  I mean early jet speeds in the 500 mph range.  According to this site https://www.fighter-planes.com/jetmach1.htm the speed of sound is 706.9 mph at 20,000 feet.  85% of that is 601mph.  I'm referring to planes slower than 600mph.

@DR01D I'll defer to Peter's engineering prowess here, but the prop tip speed problem still applies.  Careful engineering would be required, and I'm not certain a scaled-down XB-42 could hold the engines required.  You could almost certainly find a better set of props today, though...  If you'd like to reply, move us to chat, please.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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What is the bank angle that should be sustained to move in a uniform circular motion?
I want to deduce the rate of turn of targets that move in a uniform circular motion. The radius of the different circles are known in advance. I have searched for my question and found the following formulas to help me:
$$R=\frac{V^2}{11.26\tan\theta}$$
$$\omega=\frac{1,091\tan\theta}{V}$$
The variables used are:

$V$ = true airspeed in knots
$R$ = turning radius in feet
$\theta$ = bank angle in degrees
$\omega$ = rate of turn in degrees per second

As we see from the previous equations that we can get the velocity from the first equation and then substitute in the second one to get the rate of turn. But the problem in the bank angle, how can one put a reasonable values for this angle on different radius. Of course the targets that rotates in circles with small radius will different from that moves in a large radius each one will have its own reasonable bank angle and hence its own speed that it will fly with and hence its own rate of rotation.
Note:
all that because i am trying to sample the plane movement over different circles each T (secs). So i want the rate of rotation to be reasonable to be able to get enough samples to represent that movement. Also i have read that the civilian planes is different from the military ones, so it seems to be complicated topic. Any help is appreciated, thanks in advance.

The required bank angle depends on both the speed and the radius of turn. You need to know any two of these to compute the third.

A full description of these formulas are in the FAA Pilot's Handbook on page 4-34:

I hope you also appreciate help that uses the metric system. You should. It makes the calculation much simpler.
So your question is: What bank angle $\varphi$ do I get when flying a circle of a known radius $R$?
All by itself, the question cannot be answered. You need to specify a speed, either the flight speed $v$ or the angular velocity $\omega$, in order to find the proper bank angle. Then it is easy:
$$tan\varphi = \frac{v^2}{R\cdot g}\;\; \text{or}\;\;\varphi = arctan\left(\frac{v^2}{R\cdot g}\right)$$
$$tan\varphi = \frac{\omega\cdot v}{g}\;\; \text{or}\;\;\varphi = arctan\left(\frac{\omega\cdot v}{g}\right)$$
where $g$ is the gravitational acceleration and all parameters should be inserted in their SI units. $\varphi$ will be in radians!
If you don't know either speed, try some values until you get a reasonable angle, which would be not more than 30° for civilian and 75° and more for military aircraft. You can find an upper bound for that angle if you know the maximum sustained load factor of the respective plane because the roll angle and the load factor $n_z$ in a turn are directly related:
$$tan\varphi = \sqrt{n_z^2 - 1}\;\; \text{or}\;\;\varphi = arctan\left(\sqrt{n_z^2 - 1}\right)$$

to be sure that i received it well, the units of v is m/s for  r is meter for g is m/(s*s) and for bank angle is degree and for w is degree/sec. am i right?

@Ahmed: Yes for length and time, no for angles: Please use radians for $\varphi$ and rad/s for $\omega$. $\omega$ = 1 means 2$\pi$ (6.283) seconds per full circle and a bank angle of 30° is 0.5236 rad.


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What are the consequences of hail damage to a small GA plane?
I recently found a Cessna 172 that had been totaled by the insurance company because of hail damage.
The PDF from that page has detailed pictures which show hail-divots all over the the wings, body and empennage.

Can the plane still fly like this?
What effect is this likely to have on the flight characteristic of the plane?

An interesting information would be whether these planes had to land (and so where able to fly), or the damages took place on the ground. Also are you interested in "can fly" only, or also in "allowed to fly"?

That looks to me to be cosmetic. If a mechanic will sign an annual then you can fly it. I'd check to see if Cessna has any SBs out for hail damage first, though.

Of course, there is the note on the report: "PHOTOS DO NOT CLEARLY DEPICT THE HAIL DAMAGES."  So I suspect it is signficantly worse than it appears

Like anything in aviation, it depends...
There are a lot of discussions on the internet about it you can find
Here (mooney)
Here (cessna)
And Here
It is a case by case situation and depends on the extent of the hail damage and what/where the damage occurred. In many cases the issue crops up that you can not simply bang the dents out you would need to reskin the panel. This can, depending on the aircraft be a costly process.
There are planes out there flying with light hail damage and it causes no problems, in many cases its purely cosmetic. A lot of planes flying out there have other similar damage (tool drops, bird strikes, impacts from rocks etc) and fly just fine.
An good A&P would most likely be able to tell you if the particular Cessna in question is still airworthy.

@PHChilly:s answer suggested that hail damage can have adverse effects on aerodynamics just like icing. Has someone made research on this topic in wind tunnels? I would guess this is quite critical knowledge.

I dont know of any research but if you follow the links here many people seem to agree that it largely depends and should be handled on a case by case basis. Icing substantially changes the shape of the wing and GREATLY effects CG. Hail damage does not necessarily change the wing shape that drastically and does not impact CG.

@Communisty I imagine that would really also depend on the aircraft and its performance capabilities.

I have bought, flew, and sold a couple of hail damaged planes, and adjusted claims on many more.  The plane doesn't know it has hail damage and flies as it should.  Good prior point regarding structural damage, that would be very bad.  Dime to silver dollar size dents are no big deal.  The price future sale is a bigger issue.  Good luck, flyingadjuster

My plane was hail dinged up back in the 70s 2 owners before me.  Found out just how much when it was repainted - lot of red bondo filling the dings and smoothed out nice. Paint shop removed all that - only surface without dents was the ailerons, they think those were either reskinned or replaced.  Unfortunately log books were lost after that, no data. Everything else had been filled, even the stabilator. They stripped it all, cleaned any surface corrosion, acid etched it, alodyned it, filled the dents with new lightweight filler (microballoons/epoxy mix), and will be priming and painting soon.

I can tell you that hail damage does not have a significant effect on low performance aircraft such as metalized c 120s through cabin class twins unless you have structural impacts at seems in the skin. Frankly you would be hard pressed to tell the difference behind the wheel and yes even with it iced up. Your more likely to find departures from designe and test performance from poorly riged aircraft that are cross controlled before its even untied.

Airflow over the wing would be so disrupted that it'd be unsafe to fly, assuming it got off the ground to begin with. It might even be able to take off, but the critical speeds and handling qualities would be so far out of design conditions that you'd have no way of knowing what the airplane would do next.

Any sources / references?   I see many online forums where posters seem to claim it is only a cosmetic issue, and doesn't really affect handling?

Private Pilot 101. Same reason you're not supposed to take off with ice on the wing - any significant change in surface texture disrupts airflow, thus affecting lift, thus changing performance and handling characteristics. From the FAA Pilot Handbook of Aeronautical Knowledge (https://www.faa.gov/regulations_policies/handbooks_manuals/aviation/phak/media/07_phak_ch5.pdf): "As little as
0.8 millimeter of ice on the upper wing surface increases drag
and reduces aircraft lift by 25 percent." A wing that looks like somebody took a ball-peen hammer to it is going to have similar problems.

You'd be quite the test pilot going flying in an aircraft looking like that... no idea what any of your speeds are (other than, equal to or higher than what's in the POH for stall speed, equal to or lower than the POH for VNE, etc), no idea what sort of flutter issues you might encounter, no idea of a LOT of what's out there.  If the alternative were certain death from flowing lava or the advancing Mongol hoard, it might be worth trying, but for anything less, there's just no way. Far too many opportunities for things to go seriously wrong.

I want to see the data supporting the divot = ice claim. VGs significantly change the surface and improve performance. Your blanket statement, like most blanket statements, is questionsble. Airworthiness cannot be determined from a few photos.

He asked about the likely changes to flight characteristics.Comparing random dimpling from hail damage to carefully placed vortex generators is similarly a blanket statement. Given the extent of the damage pictured, I think it's reasonable to use ice for comparison.

I disagree.  The images shown, while ugly (and taken at at angle to highlight divots, btw) cannot really be compared to icing.

I don't understand the downvotes. While I think PHChilly might be a bit on the extreme side, I have to agree with the premise. Icing adds significant weight in addition to the disruption of airflow, so there are some additional factors that make those difficult to compare, but any significant change to wing shape will have unknown consequences. I'd be less concerned with it getting off the ground than I would be with what happens if/when it encounters icing again. Or even worse, what happens when it does stall? I wouldn't want to find out that its new spin characteristics are a bit extreme.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What are the rules for Public-Use Operations?
On another aviation board I'm on, (link), one poster is making the claim that operators of airplanes for Public Use Operations (eg. Police planes, Search-and-rescue aircraft, most government planes), do not even technically need a pilot's license, and do not have to meet FAA maintenance rules.
He even cites an FAA Advisory.
I find this extremely hard to believe. But I haven't found anything to refute this.
What are the rules for Public Use Operations, and what role does the FAA play in keeping them safe?

I see in the document it keeps saying something to the effect "The FAA has limited oversight of PAO, though such operations must continue to comply with the regulations applicable to all aircraft operating in the NAS. The government entity conducting the PAO is responsible for oversight of the operation, including aircraft airworthiness and any operational requirements imposed by the government entity. The government agency contracting for the service assumes the responsibility for oversight of a PAO." Which seems to say that they still need to play by the rules, but FAA may not enforce.

Part of the discussion centers around the difference in privileges afforded to private vs commercial pilots, and whether private pilots can fly Public Use aircraft. In other words, I interpret some of the discussion on certificated pilots to by type of certificate, not whether the individually holds any pilot certificate.

Short answer: public (federal or state government) aircraft operations can be exempt - but aren't always - from most FAA regulations, provided they're non-commercial operations. Any regulation that specifically applies to a "civil aircraft" doesn't apply to a public aircraft operation.

First, some definitions from 14 CFR 1.1:

Civil aircraft means aircraft other than public aircraft.
Public aircraft means any of the following aircraft [...]

The definition of "public aircraft" basically comes down to 'any aircraft the government owns or has a long-term lease on'. That potentially covers everything from Air Force One down to the Podunk Sheriff Department's C172. With that in mind, a lot of regulations apply only to civil aircraft, e.g. 14 CFR 61.3 requires a pilot's certificate to operate an aircraft (emphasis mine):

(a) Required pilot certificate for operating a civil aircraft of the
United States. No person may serve as a required pilot flight
crewmember of a civil aircraft of the United States, unless that
person:

So 61.3 doesn't apply to public aircraft. Similarly, 91.203 requires an airworthiness certificate for civil aircraft only (emphasis mine):

[...] no person may operate a civil aircraft unless it has within it the following:
(1) An appropriate and current airworthiness certificate

As for maintenance, part 43 only applies if an aircraft has that certificate, per 43.1:

this part prescribes rules governing the maintenance, preventive maintenance, rebuilding, and alteration of any—
(1) Aircraft having a U.S. airworthiness certificate;

In other words, if your aircraft is public use then you don't need a pilot's certificate to fly it, you don't need an airworthiness certificate, and therefore you don't need maintenance! See also 91.7, 91.9 and many other regulations that only apply to "civil aircraft".
Whether or not a specific flight is a public aircraft operation (PAO) or not is explained in AC 00-1.1A, Public Aircraft Operations. Pages 10 and 11 have flowcharts for determining whether a flight is a PAO but to simplify, if the aircraft is owned by the government and the flight isn't for commercial purposes (also defined in the AC) then it's a PAO. And that means many regulations don't apply to them:

What Oversight of PAO Does the FAA Have? The FAA has limited oversight of PAO,
though such operations must continue to comply with the regulations applicable to all aircraft
operating in the NAS.

Again, if you read the regulations with the civil/public distinction in mind then you can see a lot of exemptions. For example, 91.131 is the regulation for operations in class B airspace. It says (emphasis mine):

(a) Operating rules. No person may operate an aircraft within a Class B airspace area except in compliance with §91.129 and the following rules:
(1) The operator must receive an ATC clearance from the ATC facility having jurisdiction for that area before operating an aircraft in that area.
[...]
(b) Pilot requirements. (1) No person may take off or land a civil aircraft at an airport within a Class B airspace area or operate a civil aircraft within a Class B airspace area unless—
(i) The pilot in command holds at least a private pilot certificate;

That means (a) applies to all flights but (b) applies only to civil flights, i.e. a PAO flight still needs a clearance for class B airspace, but the pilot doesn't need to have a pilot's certificate. It looks to me like the regulations are generally written so that everyone has to follow the same rules of the air (for obvious safety reasons) but public aircraft are often exempt from licensing, airworthiness and maintenance requirements.
Does this all mean that US airspace is full of unlicensed pilots flying non-airworthy aircraft that haven't been maintained properly, all under the cover of government business? Not really; the government has as much interest in safe and productive flying as anyone else, and that requires qualified pilots and well-maintained aircraft. There are some plausible reasons for the PAO exemptions, like allowing military pilots to operate civil flights, and federal and state governments are often exempt from their own legislation in all sorts of areas, not just aviation.

Where it gets messy is when you have an aircraft that has civil registration/airworthiness cert but is performing public operations outside the scope/boundaries of the FARs (Omega's tankers come to mind here -- N707MQ's one of them).

"you don't need an airworthiness certificate, and therefore you don't need maintenance!" - if only regulations could bend reality like that.

@immibis Yes, it's a really extreme example of "legal is not safe" :-)

Not sure if this fully answers your question, but I think they're misreading the circular. The misread here is the word operator which does not mean pilot. The circular is saying you don't need an operator's certificate, nor necessarily comply with the requirements under part 135. Not whether the person flying the aircraft (which may or may not be the same) has to have a license. A pilot can be an operator as well, but they are two separate things.
It does, however, waive the part 91, § 91.7 airworthiness requirements and places that responsibilty with the government entity conducting the flight.

Edit: Although the circular quoted does not pertain to pilot requirements, I must defer to Pondlife's answer regarding that. In the appendices of this report .pdf the NTSB does specify that part 61.5 doesn't apply, so no licence is required by the FAA.

I'm a contractor flying public use aircraft that the Army owns. The contractors that operate these aircraft are governed by the Government organization that has contracted them and I guarantee you at a minimum they are required to have at least a commercial license with an instrument rating and have a couple thousand hours weather having a license is a requirement or not. The military has its own systems for assuring airworthiness which the contractors are contractually required to follow. Unfortunately most of the managers in these contracting companies dont realize that they dont have to follow FAA maintenance requirements and they end up following a bastardized system that is a combination of both military and FAA. This has a tendency to confuse both the prior military maintainers and the A&P mechanics that only know FAA. This dilemma is partially because the parts on the aircraft that have a civilian counter part, like the UH72 and the EC145, have to be maintained IAW FAA directives for the components requiring rebuild to go back into the supply system.

I've had public use jet transport pilots tell me that they do not have to comply with FAA requests (demands) for a "ramp check," wherein pilots of the same aircraft type in mainstream commercial operations would need to allow physical access and would have to produce requested documents demonstrating pilot, operator and aircraft compliance with requirements for designated operations. Nice option if it's true.

Welcome to Av.SE! Thank you for the personal experience angle on this. The implications of FAA jurisdiction play out in many ways. Please do stick around & contribute on various threads!

Also don't forget State Laws. For example North Carolina Chapter 63 Aeronautics, 63-20 requires all aircraft pilots to be Federally Licensed.....the loophole is that no NC State employed pilot has a License since the FAA only issues Pilot Certificates...but we are all Federally Certificated :-) It's all about the language!

This doesn't seem to answer the stated question.


text:
When should pilots observe a "Sterile Cockpit"?
A "Sterile Cockpit" means avoiding all conversations and actions not related to flight during important flight times.  (eg. No extraneous conversations, no non-important electronics, no eating, etc)
Commercial pilots are told to observe Sterile Cockpit below 10,000 feet.
But what about Private Pilots who may rarely ever go above 10,000?  When do you begin observing sterile cockpit?  When do you consider it safe to talk with passengers, take a snack, or fiddle with the new GPS?

The "sterile cockpit rule" is an informal name for 14 CFR 121.542 and 14 CFR 135.100.  It says that no non-essential duties may be done during a critical phase of flight which is defined as "all ground operations involving taxi, takeoff and landing, and all other flight operations conducted below 10,000 feet, except cruise flight".  In your example, once they level off at cruise, even if it is below 10,000 feet, it no longer applies.
While this doesn't apply to Part 91 operations like you describe, it is a very good idea to follow the rule anyway to prevent distractions and increase the level of safety.
As far as the GPS, I would recommend fiddling with it on the ground before you takeoff and be familiar with its operation before you need to use it.  Being heads down for an extended period of time is never a good idea (unless you have a safety pilot along to keep an eye on things and look for traffic)!
The full text of the regulation is:

§121.542   Flight crewmember duties.
(a) No certificate holder shall require, nor may any flight crewmember
  perform, any duties during a critical phase of flight except those
  duties required for the safe operation of the aircraft. Duties such as
  company required calls made for such nonsafety related purposes as
  ordering galley supplies and confirming passenger connections,
  announcements made to passengers promoting the air carrier or pointing
  out sights of interest, and filling out company payroll and related
  records are not required for the safe operation of the aircraft.
(b) No flight crewmember may engage in, nor may any pilot in command
  permit, any activity during a critical phase of flight which could
  distract any flight crewmember from the performance of his or her
  duties or which could interfere in any way with the proper conduct of
  those duties. Activities such as eating meals, engaging in
  nonessential conversations within the cockpit and nonessential
  communications between the cabin and cockpit crews, and reading
  publications not related to the proper conduct of the flight are not
  required for the safe operation of the aircraft.
(c) For the purposes of this section, critical phases of flight
  includes all ground operations involving taxi, takeoff and landing,
  and all other flight operations conducted below 10,000 feet, except
  cruise flight.
Note: Taxi is defined as “movement of an airplane under its own power
  on the surface of an airport.”

NASA has a great article called The Sterile Cockpit which goes into a lot more information, and includes some interesting statistics from the ASRS database about accidents/incidents that were in some way caused by not following this regulation:

48% were altitude deviations
14% were course deviations
14% were runway transgressions
14% were general distractions with no specific adverse consequences
8% involved takeoffs or landings without clearance
2% involved near mid-air collisions due to inattention and
  distractions.

Abiding by the sterile cockpit rule can help you to avoid situations like that! The article also gives some examples from the database and talks about a couple of crashes that were attributed to lack of compliance.

Good answer with sources cited. But personally, I consider that too loose.   I'm still in cruise flight until shortly before I enter the pattern for landing!   I personally tend to move to Sterile Cockpit around 5 miles outside a class-D space, when I'm first calling a class-D tower and receiving instructions.

@abelenky Yeah, the NASA article that I just added mentions that, and there is one mention of choosing a DME distance to comply.

@abelenky By the way, I would say to start observing a sterile cockpit when you start your descent, which is normally a little further out than pattern entry.  If you were cruising at pattern altitude (seems a little low to me!), then it would have to be earlier (maybe when you get the ATIS or start talking to approach).


text:
Is the Common Route, either of a SID or a STAR, related to the Runway Transition (climb phase) or the Enroute Transition (enroute phase)?
Based on this question ("What are the 'parts' of a SID procedure?"), I would like to ask for clarification concerning the role of the Common Route in a SID.
As far as I have understood, the most important part of the SID (the "runway transition" as it is called in the ARINC424 specification manual) is visualized with bold lines on the related charts. During this phase, the aircraft performs an obligatory climb with a minimum climb gradient of 200 ft/NM, unless otherwise specified. The remaning routes (common route and enroute transitions) are visualized with thinner lines and their respective Minimum Enroute Altitudes are also depicted.
What I find confusing is that the Common Route is sometimes visualized with bold lines and without Minimum Enroute Altitudes and other times with thinner lines and with their respective MEAs.
An example can be found if one compares FINZZ2 with EPKEE:

Is there a reason for having the Common Route sometimes as part of the Runway Transition (without MEAs) and other times as part of the Enroute Transitions (with MEAs)?
EDIT:
Τhe same distinction also applies to STARs, where common route can either be part of enroute or runway transitions. An example can be found here and here (OHSEA TWO Arrival, KSNA). Notice that common route is visualized with bold lines, same as runway transitions. However, most of the arrival procedures, as far as I can tell, use it as part of the enroute phase (containing MEA, drawn with thinner lines etc.).

Obstacle or terrain clearance probably?

@MichaelHall Thank you for your input. It seems reasonable but in this case why not just end the Runway Transition at the waypoint which provides the desired obstacle clearance, instead of renaming the remaining path as "common route"? There is already a common path which is part of the runway transition (APUUU TO EPKEE). Why ending runway transition at EPKEE and deciding that the rest will be the Common Route? I know I am probably overthinking this, but I am really curious if there is any reason behind it. Notice that there is no way of coding this distinction with ARINC424.

Is it possible that what you are observing is simply normal "process variation" where different designers working on different areas to a common design guideline just come up with a slightly different way of doing things?  If either way of depicting this is permissible then does it really matter?

What airports are these SIDs from? Your links are going to break every time the FAA makes a website change.

This took some digging.
There seems to be some mixing of the terms in the question, using different publications, mixing between STARS and SIDs, and maybe neglecting to read the textual route descriptions. The following is only about SIDs.
SIDs, unlike obstacle departure procedures, are primarily developed to assist ATC with traffic management. They get you up and away from other airport traffic.

There are two types of DPs; those developed to assist pilots in
obstruction avoidance (referred to as ODP) and those developed to
communicate air traffic control (ATC) clearances (referred to as SID).
FAA 8260.46j

According to the FAA chart legend, FAA chart user guide, and FAA charts. The thick lines in the FAA charts are the "departure route" the thin lines are the "transition routes".
There is no "common route" or "runway transition" for a SID these terms are only used for STARs (searching 8260.3E)
Transition routes are not part of the SID, they are engineered much like mainline en-route airways. (V J T Q routes)
The SID ends where the bold lines end, which is also where the detailed textual description ends. Normally this ending fix is also the name of the SID and it is the generic transition fix from which you can file to some other en-route fix in your flight plan such as FINZZ3.FINZZ ALBAS V25 MZB
EPKEE7 (and EPKEE6) seems to be an exception, but I found this in 8260.46j

Design SIDs to terminate at a fix/NAVAID depicted on an IFR en route
chart, at an altitude that will allow random IFR flight, or at a
position and altitude where ATC radar service is provided.

Which likely fits EPKEE
Along with

When ATC requests an altitude restriction for a fix located on a
transition route, it must be at or above the specified minimum en
route altitude (MEA) for the route (see note in paragraph
2-1-1.e(2)(b)). Do not raise an MEA to support ATC operational
requirements; use fix crossing altitudes where operationally needed.

There are also some guidelines on how departures should be worded without excess complexity, to reduce pilot workload.
The textual description of EPKEE7 says to maintain fl230 on transitions unless filed lower, so the SID section from EPKEE to POIZN is a way of ensuring at least 15000 between EPKEE and POIZN for ATC purposes. (APUUU to POIZN gives clearance over some V routes)
The OROCA for the EPKEE-POIZN is 9100 with several victor MEAs of 7000, so making this a transition route would end up with a low MEA. Also the transition routes after POIZN go over a corner of an MOA. So it is apparently just a matter of altitude control.
Especially as this is a jet-only RNAV SID and the only practical time those transitions would really be used is with a radio failure.


text:
Which Navigation Specifications support an RF leg?
I am studying FAA's rules concerning navigation specifications and the utilization of RF legs in an Instrument Flight Procedure. However, I am having trouble understanding if an RF leg is supported by RNAV procedures (RNAV 1 specifically) or only by RNP procedures. After some search in FAA's IFP coded database (CIFP file), I realised that only SIDs labeled as RNP 1 support RF legs, at least as far as FAA's procedures are concerned. However, an introductory page on FAA's site, concerning RNAV and PBN procedures, includes the RF as a general RNAV leg type.
Am I missing something? Could someone point me in the right direction?
EDIT: I think the question that pretty much sums up my confusion is if it's possible for a SID or STAR with only an "RNAV 1", instead of "RNP 1", requirement to contain an RF leg. Because so far I have only encountered RNP 1 SIDs with RF legs, such as STAYY ONE

First is understanding the difference between RNAV and RNP.  This related question: What is the difference between RNAV and RNP? provides a basic answer to that part of your question.
To your specific question, RNAV procedures predate the acceptance of RNP standards. Since RNP standards have been established as the basis of navigation, you should expect that all procedures published in the future would be RNP procedures.
As a reference, I can point to RTCA Document DO-236C, Minimum Aviation System Performance Standards: Required Navigation Performance for Area Navigation (MASPS). This document describes how RNP standards will be applied in the airspace. That covers airspace design, procedures and equipment used to fly within RNP airspace. Performance Based Navigation (PBN) adds aircraft specific performance requirements to the RNP concept.
Definitions from DO-236C which references ICAO Doc 9613:

RNAV system (as defined by ICAO Doc 9613): A navigation system which
  permits aircraft operation on any desired flight path within the
  coverage of station-referenced NAVAIDs or within the limits of the
  capability of self-contained aids, or a combination of these. An RNAV
  system may be included as part of a flight management system (FMS).
RNP system (as defined by ICAO Doc 9613): An area navigation system
  which supports onboard performance monitoring and alerting. An area
  navigation system capable of achieving the performance requirement of
  an RNP specification is referred to as an RNP system.
PBN (as defined by ICAO Doc 9613): Area navigation based on
  performance requirements for aircraft operating along an ATS route, on
  an instrument approach procedure or in a designated airspace.

Also from DO-236C, section 3.2 Path Definition Requirements, we find:

3.2.1.1 Navigation Database Flight Path Legs for Terminal Area Procedures
Permissible leg types for the definitions of the RNP flight paths for
  all navigation database procedures (e.g. SIDs, STARS, Approaches
  including missed approaches), with the exception of airways, are
  limited to the following:
Permissible Leg Types          ARINC 424 Leg Type
- Track to Fix leg                      TF
- Radius to Fix leg                     RF
- Initial Fix                           IF
- Hold legs                        HM, HA and HF
- Fix to Altitude leg                   FA
- Direct to Fix leg                     DF
- Course to Fix leg                     CF
Note 1: RNP procedures that require the repeatability and predictability of
a specified ground track should be constructed using IF, TF, and RF legs only.
Note 2: If the FA, DF and CF types are used, they may introduce less
repeatability and predictability.

As you can see from the above, RNP procedures are built using RNAV leg types. RF legs are an RNAV leg type unlike the similar AF leg which is a DME arc that can be flown with a DME.
Realistically, you will only find RF legs within SIDs, STARs, and IAPs as that's where they are needed. They are useful when you need a large change of direction which is common in the terminal environment but not so much while enroute.

@StephenS  you are correct.  I should have included approaches in my statement. RF legs are a common part of PBN approach procedures.

@Gerry I really appreciate the answer, helped me to clear up some things. But since I would like to make sure that I have understood this, could you clarify one more thing: Is it possible for a SID or STAR with only a "RNAV 1 required", instead of "RNP 1 required", to contain an RF leg? Because so far I have only encountered RNP 1 SIDs with RF legs, such as STAYY ONE

I don't think it is. All the procedures I see that say "RNAV 1" include the note DME/DME/IRS or GPS required.  That implies older FMSes that aren't up to the RNP standard. Most FMS of that era do not support RF legs. Even in the current RNP universe, RF legs are still not all that common.


text:
Does a Navigation Specification characterize all legs of a terminal procedure?
I am studying the different parts of Instrument Flight Procedures, as they are described by FAA's manuals, such as the Aeronautical Information Manual, TERPS and Performance Based Navigation. For example, during the transition from the feeder phase of an Instrument Approach Procedure to the initial phase, I noticed that under certain conditions the procedure's navigation specification can be changed. For example, in RNAV GPS IAPs, feeder routes can be RNAV 1, while the subsequent initial and intermediate phases switch to an RNP APPROACH specification. Another example is during a SID or STAR procedure, where the navigation specification changes when the route crosses the circle defined by the associated airport and a 30 nautical miles radius.

TERPS 1-8 2.b STAR/feeder and approach legs designed to cross within 30 NM of the ARP more than once do not change to RNAV 1/RNP 1/A-RNP (terminal mode) until the 30 NM limit is crossed for the last time. A departure or missed approach segment designed to cross a point 30 NM from the ARP more than once changes to RNAV 2/RNP 2/A-RNP (en route mode) and remains expanded when it crosses the boundary the first time.

Having said all that, could someone help me to understand under which circumstances it is possible for the navigation specification of a procedure to change and, particularly, if it is more accurate to associate a navigation specification with each leg of a procedure, rather than having one navigation specification that characterizes the whole route.